Help calculating a series I understand the process in calculating a simple series like this one
$$\sum_{n=0}^4 2n$$
but I do not understand the steps to calculate a series like this one
$$\sum_{n=1}^x n^2$$
I have an awesome calculator and know how to use it, so I know the solution
$$\frac{x^3}{3} + \frac{x^2}{2} + \frac{x}{6}$$
but I don't know the steps.  All the explanations I find (like those on Purple Math) are for summations like the first one I listed.  If someone can provide a detailed explanation on how to calculate a summation like mine or provide a link to one it would be much appreciated.
 A: $$(k+1)^3-k^3 = 3k^2 + 3k + 1$$ Now sum it up from $k=1$ to $n$. Notice the telescopic summation on the left side and use $\displaystyle \sum_{k=1}^n k = \frac{n(n+1)}{2}$ to get the answer. This is a standard technique to compute such sums.
A: Yes, you can use differences to find such formulas. Many times there are also easier ways (involving tricks). For this one you can use that
$$\sum_{i=0}^n i^3 = \left ( \sum_{i=0}^n (i+1)^3 \right ) - (n+1)^3$$
From here it is an easy computation to find the answer:
$$\sum_{i=0}^n i^3 = \sum_{i=0}^n i^3 + \sum_{i=0}^n 3i^2 + \sum_{i=0}^n 3i + \sum_{i=0}^n 1 -(n+1)^3$$
which gives
$$\sum_{i=0}^n i^2 = \frac{-\frac{3n(n+1)}{2} - n -1 + n^3 +3n^2 +3n +1}{3} $$
$$\sum_{i=0}^n i^2 = \frac{n^3}{3} + \frac{n^2}{2} + \frac{n}{6}$$
A: Another approach is to know (guess) that $\sum_{n=1}^xn^2$ is cubic in $x$.  Similar to an integral, the sum adds one degree.  Then you can just say $\sum_{n=1}^xn^2=Ax^3+Bx^2+Cx+D$.  If you calculate that the sum up to 1,2,3,4 is 1,5,14,30 (or start with 0 and sum to 0 makes it a bit easier) you can just solve the simultaneous equations for A,B,C, and D.
